USING CRAMERS RULE TO SOLVE SYSTEMS OF EQUATIONS WITH 3 VARIABLES

Problem 1 :

-x + y = -5

-5x - 2y + 6z = -5

-4x - y + 2z = 8

Solution:

Δ = -1(-4 + 6) - 1(-10 + 24) + 0

= -1(2) - 1(14)

= -2 - 14

Δ = -16

Δ₁ = -5(-4 + 6) - 1(10 - 48) + 0

= -5(2) - 1(-58)

= -10 + 58

Δ₁ = 48

Δ₂ = -1(-10 - 48) + 5(-10 + 24) + 0

= -1(-58) + 5(14)

= 58 + 70

Δ₂ = 128

Δ₃ = -1(-16 - 5) - 1(-40 - 20) - 5(5 - 8)

= -1(-21) - 1(-60) - 5(-3)

= 21 + 60 + 15

Δ₃ = 96

By Cramer's rule,

So, the values of x, y and z are -3, -8 and -6 respectively.

Problem 2 :

-4x - 2y - z = -11

-x - 2y = -6

x - y - 5z = 5

Solution:

Δ = -4(10 + 0) + 2(5 - 0) - 1(1 + 2)

= -4(10) + 2(5) - 1(3)

= -40 + 10 - 3

Δ = -33

Δ₁ = -11(10 + 0) + 2(30 - 0) - 1(6 + 10)

= -11(10) + 2(30) - 1(16)

= -110 + 60 - 16

Δ₁ = -66

Δ₂ = -4(30 - 0) + 11(5 - 0) - 1(-5 + 6)

= -4(30) + 11(5) - 1(1)

= -120 + 55 - 1

Δ₂ = -66

Δ₃ = -4(-10 - 6) + 2(-5 + 6) - 11(1 + 2)

= -4(-16) + 2(1) - 11(3)

= 64 + 2 - 33

Δ₃ = 33

By Cramer's rule,

So, the values of x, y and z are 2, 2 and -1 respectively.

Problem 3 :

5x + 2y + 2z = 9

-6x - 4y - 3z = -19

x - 2y = -9

Solution:

Δ = 5(0 - 6) - 2(0 + 3) + 2(12 + 4)

= 5(-6) - 2(3) + 2(16)

= -30 - 6 + 32

Δ = -4

Δ₁ = 9(0 - 6) - 2(0 - 27) + 2(38 - 36)

= 9(-6) - 2(-27) + 2(2)

= -54 + 54 + 4

Δ₁ = 4

Δ₂ = 5(0 - 27) - 9(0 + 3) + 2(54 + 19)

= 5(-27) - 9(3) + 2(73)

= -135 - 27 + 146

Δ₂ = -16

Δ₃ = 5(36 - 38) - 2(54 + 19) + 9(12 + 4)

= 5(-2) - 2(73) + 9(16)

= -10 - 146 + 144

Δ₃ = -12

By Cramer's rule,

So, the values of x, y and z are -1, 4 and 3 respectively.

Problem 4 :

4a + 4c = 4

4a - 3b + c = -14

-2a - 3b - 5c = -20

Solution:

Δ = 4(15 + 3) - 0 + 4(-12 - 6)

= 4(18) + 4(-18)

= 72 - 72 

Δ = 0

Δ₁ = 4(15 + 3) - 0 + 4(42 - 60)

= 4(18) + 4(-18)

= 72 - 72

Δ₁ = 0

Δ₂ = 4(70 + 20) - 4(-20 + 2) + 4(-80 - 28)

= 4(90) - 4(-18) + 4(-108)

= 360 + 72 - 432

Δ₂ = 0

Δ₃ = 4(60 - 42) - 0 + 4(-12 - 6)

= 4(18) + 4(-18)

= 72 - 72

Δ₃ = 0

Since Δ = 0, Δ₁ = 0, Δ₂ = 0 and Δ₃ = 0 and atleast one of the element in Δ is non zero. 

Then the system is consistent and it has infinitely many solution.

Problem 5 :

4x - 4y + 2z = -14

4x + 2y = 14

-3y + z = -10

Solution:

Δ = 4(2 + 0) + 4(4 - 0) + 2(-12 - 0)

= 4(2) + 4(4) + 2(-12)

= 8 + 16 - 24

Δ = 0

Δ₁ = -14(2 + 0) + 4(14 + 0) + 2(-42 + 20)

= -14(2) + 4(14) + 2(-22)

= -28 + 56 - 44

Δ₁ = -16 ≠ 0

Here Δ = 0 but Δ₁ ≠ 0.

So, the system is inconsistent and it has no solution.